Question

`lim_(n->oo)prod_(k=2)^n(1-1/k^2)`?

Answer 1

`prod_(k=2)^oo (1-1/k^2) = 1/2`

Explanation:

Consider the sequence of integrals:

`I_m = int_0^(pi/2) sin^mx dx`

As the integrand is defined and continuous in the whole interval, these integrals exist for every `m in NN`.

Evaluate now the integral by parts:

`I_m = -int_0^(pi/2) sin^(m-1)x d(cosx)`

`I_m = -[sin^(m-1)x cosx]_0^(pi/2) + int_0^(pi/2) (m-1)sin^(m-2)xcos^2xdx`

`I_m = (m-1) int_0^(pi/2) sin^(m-2)x(1-sin^2x)dx`

`I_m = (m-1) I_(m-2) - (m-1)I_m`

and finally we have the recursive formula:

`I_m = (m-1)/m I_(m-2)`

so that:

`I_(m-2)/I_m= m/(m-1)`

and:

`lim_(m->oo) I_(m-2)/I_m= 1`

Given that:

`I_0 = int_0^(pi/2) dx = pi/2`

`I_2 = int_0^(pi/2) sin^2xdx = pi/4`

consider now for every `m` the ratio:

`I_(m-1)/I_(m+1) = I_(m-1)/I_m I_m/I_(m+1)`

and evaluate each ratio using the recursive formula:

`I_(m-1)/I_(m) = ((m-2)/(m-1) I_(m-3))/((m-1)/(m) I_(m-2)`

`I_(m-1)/I_(m)= ((m(m-2))/(m-1)^2 ) I_(m-3)/I_(m-2)`

`I_(m-1)/I_(m) = (((m-1+1)(m-1-1))/(m-1)^2 ) I_(m-3)/I_(m-2)`

`I_(m-1)/I_(m) = (((m-1)^2-1)/(m-1)^2 ) I_(m-3)/I_(m-2)`

`I_(m-1)/I_(m) = (1-1/(m-1)^2 ) I_(m-3)/I_(m-2)`

and:

`I_(m)/I_(m+1)= ((m-1)/(m) I_(m-2))/((m)/(m+1) I_(m-1)`

`I_(m)/I_(m+1)= (((m+1)(m-1))/m^2) I_(m-2)/ I_(m-1)`

`I_(m)/I_(m+1)= ((m^2-1)/m^2) I_(m-2)/ I_(m-1)`

`I_(m)/I_(m+1)= (1-1/m^2) I_(m-2)/ I_(m-1)`

Then:

`I_(m-1)/I_(m+1) = (1-1/m^2) I_(m-2)/ I_(m-1)(1-1/(m-1)^2 ) I_(m-3)/I_(m-2)`

`I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 ) I_(m-3)/I_(m-1)`

and iterating:

`I_(m-1)/I_(m+1) = (1-1/m^2)(1-1/(m-1)^2 )(1-1/(m-2)^2)(1-1/(m-3)^2 ) I_(m-5)/I_(m-3)`

or:

`I_(m-1)/I_(m+1) = prod_(j=0)^p (1-1/(m-j)^2) I_(m-p-2)/I_(m-p)`

So for `p=m-2` we have:

`I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2) I_(0)/I_(2)`

`I_(m-1)/I_(m+1) = prod_(j=0)^(m-2) (1-1/(m-j)^2)(pi/2)/(pi/4)`

`I_(m-1)/I_(m+1) = 2 prod_(j=0)^(m-2) (1-1/(m-j)^2)`

Change now the index to `k= m-j`:

`I_(m-1)/I_(m+1) = 2 prod_(k=2)^m (1-1/k^2)`

so:

`prod_(k=2)^m (1-1/k^2) = 1/2 I_(m-1)/I_(m+1)`

and passing to the limit:

`prod_(k=2)^oo (1-1/k^2) = 1/2 lim_(m->oo) I_(m-1)/I_(m+1) = 1/2`

Answer 2

`lim_(n->oo)prod_(k=2)^n(1-1/k^2)`

`= lim_(n->oo)prod_(k=2)^n ((k-1)(k+1))/k^2`

`= lim_(n->oo) underbrace( ( (1)(3))/2^2)_(k = 2) * underbrace( ((2 )(4))/3^2)_(k = 3) * underbrace( ((3)(5))/4^2)_(k = 4) * underbrace( ((4)(6))/5^2)_(k = 5) ... * underbrace( ((n-3)(n-1))/(n-2)^2)_(k = n-2) * underbrace( ((n-2)(n))/(n-1)^2)_(k = n-1) * underbrace( ((n-1)(n+1))/n^2)_(k = n)`

`= lim_(n->oo) underbrace( ( (1))/2 )_(k = 2) * underbrace( ((1 )(1))/1^2)_(k = 3) * underbrace( ((1)(1))/1^2)_(k = 4) * underbrace( underbrace( ((1)(6))/5 )_(k = 5) * ... * underbrace( ((n-3)(1))/(n-2) )_(k = n-2) )\_(" Not fully factored ") * underbrace( ((1)(1))/(1)^2)_(k = n-1) * underbrace( ((1)(n+1))/n )_(k = n)`

The pattern : every term, in conjunction with the immediately preceding and succeeding terms, reduces to `1`. This is not true for the first and last term which are as shown.

`= lim_(n->oo) 1/2 * 1 * 1 * ... * 1 * (n+1)/ n`

`= 1/2 lim_(n->oo) 1+1/n = 1/2`